1.3 maths class9 new book
Usually students feel that word problems are difficult to solve. Lets start exercise 1.3 maths class9 in an easy way. Applications of real numbers in daily life are given in ex 1.3 of new book of grade 9 math for the students of Punjab. Students will check temperature conversion, profit and loss, find the numbers, find the ages of father and son and find the length in class 9 maths chapter 1 exercise 1.3 Punjab boards. Lets start carefully.
Q1 Sum of three consecutive integers if forty-two, find the three integers.
Let the integers are x, x+1 and x+2
x+x+1+x+2 = 42
3x+3 = 42
3x = 42-3
3x = 39
x= 13
Integers are 13,13+1 and 13+2 i.e 13,14 and 15.
Q2$$\;The\;diagram\;right\;angled\;triangle\\\;ABC\;in\;which\;the\;length\;of\;AC\;is\\(\sqrt3\;+\;\sqrt5\;)\;cm.\;The\;area\;of\;triangle\\ABC\;is\;(1+\sqrt{15}\;)\;cm^2.\;Find\;the\;length\\AB\;in\;the\;form\;(a\sqrt3\;+\;b\sqrt5\;)\;cm,\\where\;a\;and\;b\;are\;integers.$$
$$Area\;of\;right\;triangle\;=\;\frac12(Base)(Perpendicular)\\\\1+\sqrt{15}=\frac12(\sqrt5\;+\;\sqrt3\;)\;(AB)\\\\\frac{2(1+\sqrt{15})}{\sqrt5\;+\;\sqrt3}=\;AB\\\\AB=\;\frac{2(1+\sqrt{15})}{\sqrt5\;+\;\sqrt3}\\\\Rationalize\;with\;\sqrt5\;-\;\sqrt3\\\\AB=\;\frac{2(1+\sqrt{15})}{\sqrt5\;+\;\sqrt3}\;\times\frac{\sqrt5\;-\;\sqrt3}{\sqrt5\;-\;\sqrt3}\\\\AB=\;\frac{2(1+\sqrt{15})\;(\sqrt5\;-\;\sqrt3)}{{(\sqrt5\;)}^2\;-\;{(\sqrt3)}^2}\\\\AB\;=\;\frac{2(\sqrt5-\sqrt3+\sqrt{75}-\sqrt{45})}{5-3}\\\\AB=\;\;\frac{2(\sqrt5-\sqrt3+\sqrt{25\times3}-\sqrt{9\times5})}2\\\\AB=\;\sqrt5-\sqrt3+5\sqrt3-3\sqrt5\\\\AB\;=4\sqrt5-\;2\sqrt5\\\\Hence,\;a=4\;,\;b=-2$$
Q3 $$\;\;A\;rec\tan gle\;has\;sides\;of\;length\\(2+\sqrt{18})\;m\;and\;(\;5-\frac4{\sqrt2})\;m.\;Epress\\the\;area\;of\;rectangle\;in\;the\;form\;of\\a+b\sqrt2,\;where\;a\;and\;b\;are\;integers$$
$$Let\\Length\;=L\;=\;(2+\sqrt{18})\;m\;\\Width\;=\;W\;=\;(\;5-\frac4{\sqrt2})\;m\\\\Area\;of\;rec\tan gle\;=\;LW\\\\Area\;of\;rec\tan gle\;=\;\;(2+\sqrt{18})\;(\;5-\frac4{\sqrt2})\\\\=10\;-\;\frac8{\sqrt2}\;+\;5\sqrt{18\;}-\;\frac{4\sqrt{18}}{\sqrt2}\\\\=\;10\;-\;\frac{8\;\sqrt2}{\sqrt2\;\sqrt2}+\;5\;\sqrt{9\times2}\;-\;4\sqrt{\frac{18}2}\\\\\\=10\;-\;\frac{8\;\sqrt2}2\;+\;5\;(3\sqrt2)\;-\;4\sqrt9\\\\=\;10-4\sqrt2+15\sqrt2\;-4(3)\\\\=10\;+11\sqrt2-12\\\\=\;-2\;+11\sqrt2\\\\Hence\;a=-2\;,\;b=11$$
Q4 Find two numbers whose sum is 68 and difference is 22
Let the numbers are x and y.
x + y = 68 and x – y = 22
Adding the above two equation, we get
x + y + x – y = 68 + 22
2x = 90
x = 45
Putting x = 45 in x + y = 68 , we get
45 + y = 68
y = 68 – 45
y = 23
Number are 45 and 23 whose sum is 68 and difference is 22.
Q5 Weather in Lahore was usually warm during the summer of 2024. The TV news reported temperature as high as 48 degree Celsius. By using the formula find the temperature as Fahrenheit scale.
$$Formula:\\F^\circ\;=\;\frac95C^\circ\;+\;32\\\\F^\circ\;=\;\;1.8(48)\;+\;32\\\\F^\circ\;=\;\;86.4\;+\;32\\\\F^\circ\;=118.4$$
Q6 Sum of ages of the father and son is 72 years. Six years ago, the father’s age was 2 times the age of the son. What was son’s age six years ago?
Let
Current age of father = F
Current age of son = S
F + S = 72 (i)
Six years ago
Age of father = F – 6
Age of son = S -6
F – 6 = 2 ( S – 6 ) or F – 6 = 2S – 12 or F – 2S = -12 + 6
F – 2S = -6 (ii)
Subtracting equation (ii) from equation (i)
F + S – ( F – 2S ) = 72 – (-6)
F + S – F + 2S = 72 + 6
3S = 78
Dividing both sides by 3
S = 26
Age of son 6 years ago = 26 – 6 = 20
Lets find age of father
Put S = 26 in equation (i)
F + 26 = 72 0r F = 72 – 26
Its mean F = 46
Putting F = 46 we get the age of father 6 years ago = 46 – 6 = 40
Q7 Mirha bought a toy for Rs 1520. What will the selling price be to get a 15% profit?
Purchased price = 1520 Rs
Profit Percentage = 15 %
Profit = 1520 ( 15 / 100 ) = 22800 / 100 = 228
Selling price = 1520 + 228 = 1748
Q8 Annual income of Tayyab is Rs 960000 while exempted amount is Rs 130000. How much tax would he have to pay at the rate of 0.75 % ?
Total amount = 960000 Rs
Exempted amount = 130000
Rate of tax = 0.75 %
Taxable amount = 960000 – 130000 = 830000
Tax = 830000 ( 0.75 % ) = 830000 ( 0.0075 ) = 6225 Rs
Q9 Find compound markup on Rs 375000 for one year at the rate of 14 % compounded annually.
Compound markup = 375000 ( 14 % ) = 375000 (0.14 ) = 52500 Rs
2nd method:-
$$A\;=\;P\;{(\;1+r\;)}^t\\A\;=\;375000\;{(\;1+\;14\;\%\;)}^1\;\;=\;375000\;(\;1+\;0.14\;)\\\;=\;375000\;(1.14)\;\;=\;427500\\$$
Markup = A – P = 427500 – 375000 = 52500 Rs
